Question

According to the MIT Airline Data Project, American Airlines controlled 15.5% of the domestic market during a recent year. A random sample of 125 domestic passengers that year was selected. Using the normal approximation to the binomial distribution, what is the probability that 10 or fewer passengers from this sample were on American Airline flights

Answer 1

Answer:

0.0143

Step-by-step explanation:

For us to get the probability of the passenger being an American, which is 15.5% = 15.55÷100

= 0.155

Which is the probability P

The sample size n is = 155

To get the probability of the person is not on the flight, which is q

Where q is 1-P

That's

q =1 - p= 0.845

P(X < A) = P(Z < (A - mean)/standard deviation)

Our mean is given as np ( population size × probability

= 125 x 0.155

= 19.375

And the standard deviation is given as = √npq

= √ (125 x 0.155x 0.845)

=√16.3719

= 4.0462

P(10 or lesser passengers that are on the American Airline flights) is given as = P(X \ 10)

=P(Z<{(A- mean) ÷standard deviation }

We have as

= P(Z < (10.5 - 19.375)/4.0462)

= P(Z < -2.19)

= 0.0143 as the probability that 10 or fewer passengers from this sample were on American Airline flights.

Answer 2

Answer:

0.0143

Step-by-step explanation:

In this question, we are asked to use the binomial distribution to calculate the probability that 10 or fewer passengers from a sample of MIT data project sample were on American airline flights.

We proceed as follows;

The probability that a passenger was an American flight is 15.5%= 15.55/100 = 0.155

Let’s call this probability p

The probability that he/she isn’t on the flight, let’s call this q

q =1 - p= 0.845

Sample size, n = 155

P(X < A) = P(Z < (A - mean)/standard deviation)

Mean = np

= 125 x 0.155

= 19.375

Standard deviation = √npq

= √ (125 x 0.155x 0.845)

= 4.0462

P(10 or fewer passengers were on American Airline flights) = P(X \leq 10)

= P(Z < (10.5 - 19.375)/4.0462)

= P(Z < -2.19)

= 0.0143